Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A belt runs on an 80 mm driving pulley at 960 rpm. The tension is 500 N on the tight side and 250 N on the slack side. Find the power P transmitted, in kW.
d1 = 80 mmn1 = 960 rpmT1 = 500 NT2 = 250 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 125 mm driving pulley at 1450 rpm to a 250 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 1200 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 125 mmn1 = 1450 rpmd2 = 250 mmC = 600 mmμ = 0.25T1 = 1200 N
P = kW
3.Torsion of shaftsMedium
A stepped solid steel shaft is fixed at A. Segment AB has diameter 25 mm and length 1.2 m; segment BC has diameter 20 mm and length 1.2 m. A torque of 100 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
T = 100 N·md1 = 25 mm, L1 = 1.2 md2 = 20 mm, L2 = 1.2 m
τmax = MPa
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 20 mm is fixed at A. A torque of 200 N·m acts at B, 1.5 m from A, and a torque of 100 N·m in the same direction acts at the free end C, a further 2.5 m along. Find the maximum shear stress τmax in the shaft.
d = 20 mmTB = 200 N·mTC = 100 N·mAB = 1.5 mBC = 2.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 80 MPa, σᵧ = 70 MPa and τₓᵧ = 20 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 80 MPaσy = 70 MPaτxy = 20 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 150 MPa, σᵧ = −70 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.
σx = 150 MPaσy = −70 MPaτxy = −50 MPaθ = 35°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 90 kNd = 10 mmL = 1.5 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaP = 60 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 70 GPa (aluminium)d = 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 85 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 85 kNn = 3L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 2 cycles later by a peak of 4.7 mm. Find the damping ratio ζ.
x0 = 10 mmx2 = 4.7 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 12 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 12 kN/mk2 = 12 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 1.005 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt speed
v = 4.021 m/s
Power
P = (500 − 250)·4.021 = 1005 W = 1.005 kW
Answer
P = 1.005 kW
2.Gears & gear trains
Answer: P = 5.918 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Angle of wrap
θ = 2.933 rad
Slack-side tension
eμθ = e0.7332 = 2.082, T₂ = 1200/2.082 = 576.4 N
Power
P = (1200 − 576.4)·9.49 = 5.918 kW
Answer
P = 5.918 kW
3.Torsion of shafts
Answer: τmax = 63.66 MPa
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Stress in each segment
T = 100 N·m = 100000 N·mm
τAB = 16·100000/(π·25³) = 32.59 MPa
τBC = 16·100000/(π·20³) = 63.66 MPa
Answer
τmax = 63.66 MPa
4.Torsion of shafts
Answer: τmax = 191 MPa
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Internal torques
Sense of TC taken as positive.
TBC = TC = 100 N·m
TAB = TC + TB = 100 + 200 = 300 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 300 N·m
τmax = 16·300000/(π·20³) = 191 MPa
Answer
τmax = 191 MPa
5.Mohr’s circle & stress transformation
Answer: θp = 37.98 °
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Angle on the circle
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·20, 80 − 70) = 75.96°
Angle on the element
θp = 37.98°
Answer
θp = 37.98°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 30.64 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Transformation equation
σθ = 40 + 110·0.342 + −50·0.9397 = 30.64 MPa
Answer
σθ = 30.64 MPa
7.Axial stress & strain
Answer: δ = 24.56 mm
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·10²/4 = 78.54 mm²
σ = P/A = 90·1000/78.54 = 1146 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 1146/70000 = 0.01637 = 16370 µε
Elongation
δ = ε·L = 0.01637·1500 mm = 24.56 mm
Answer
δ = 24.56 mm
8.Axial stress & strain
Answer: σ2 = 34.33 MPa
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Stiffnesses
k₁ = 200000·490.9/400 = 245437 N/mm
k₂ = 100000·1257/200 = 628319 N/mm
Share of segment 2
F2 = 60000·628319/873755 = 43146 N
Stress
σ2 = 43146/1257 = 34.33 MPa (compression)
Answer
σ2 = 34.33 MPa
9.Column buckling
Answer: Pcr = 5.888 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴
Effective length
KL = 2·3000 = 6000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·306796/6000² = 5888 N = 5.888 kN
Answer
Pcr = 5.888 kN
10.Column buckling
Answer: d = 74.06 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Required critical load
Pcr = n·P = 3·85000 = 255000 N
KL = 0.5·4000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·255000·2000²/(π³·70000))^(1/4) = 74.06 mm
Answer
d = 74.06 mm
Check: the slenderness KL/r is 108, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.05997
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Logarithmic decrement
δ = (1/2)·ln(10/4.7) = 0.3775
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3775/√(39.48 + 0.1425) = 0.05997
Answer
ζ = 0.05997
12.Mechanical vibrations
Answer: fn = 5.257 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 12 + 12 = 24 kN/m
keq = 24·20/(24 + 20) = 10.91 kN/m
Natural frequency
ωn = √(10909/10) = 33.03 rad/s
fn = 33.03/(2π) = 5.257 Hz
Answer
fn = 5.257 Hz
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